Question

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Data. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.

Answer 1

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Explanation:

The question is incomplete:

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.

We start calculating the sample mean M and standard deviation s:

`M=(1)/(n)\sum_(i=1)^n\,x_i\\\\\\M=(1)/(10)(18+12+20+17+14+15+13+11+21+17)\\\\\\M=(158)/(10)\\\\\\M=15.8\\\\\\`

`s=\sqrt{(1)/(n-1)\sum_(i=1)^n\,(x_i-M)^2}\\\\\\s=\sqrt{(1)/(9)((18-15.8)^2+(12-15.8)^2+(20-15.8)^2+. . . +(17-15.8)^2)}\\\\\\s=\sqrt{(101.6)/(9)}\\\\\\s=√(11.3)=3.4\\\\\\`

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=15.8.

The sample size is N=10.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

`s_M=(s)/(√(N))=(3.4)/(√(10))=(3.4)/(3.162)=1.075`

The degrees of freedom for this sample size are:

df=n-1=10-1=9

The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_M=3.25 \cdot 1.075=3.49`

Then, the lower and upper bounds of the confidence interval are:

`LL=M-t \cdot s_M = 15.8-3.49=12.3\\\\UL=M+t \cdot s_M = 15.8+3.49=19.3`

The 99% confidence interval for the mean germination time is (12.3, 19.3).